An attempt to prove Kadeishvili's theorem

Question

I was browsing various papers and encountered an interesting result, namely Kadeishvili's theorem which states the following:

Let $A^*$ be an $A_\infty$-algebra in chain complexes and let $H^*(A^*)$ be the cohomology ring of $A^*$. There is an $A_\infty$-algebra structure on $H^*(A^*)$ with $m'_1=0$ and $m'_2$ induced by the multiplication on $m_2$, constructed from the $A_\infty$-structure of $A^*$, such that there is a quasi-isomorphism of $A_\infty$-algebras $H^*(A^*)\to A^*$ lifting the identity of $H^*(A^*)$. This $A_\infty$-algebra structure on $H^*(A^*)$ is unique up to quasi-isomorphism.

(See https://ncatlab.org/nlab/show/Kadeishvili%27s+theorem for more details)

I saw a sketch of a proof but it wasn't presented in full detail, so I decided to prove the theorem myself as a little exercise of the $A_\infty$-identities. However, I have absolutely no background in $A_\infty$-algebras, so I'd like know if my attempt was correct at all...

For the sake of clarity, here is some relevant information:

1. The morphisms $m_n$: if $A^*$ is graded vector space over a field $k$, then an $A_\infty$-structure on $A^*$ is given by morphisms $m_n\colon (A^*)^{\otimes n}\to A^*$ of degree $(2-n)$, satisfying $$\sum_{r+s+t=n}(-1)^{r + st}m_{r+1+t}(1^{\otimes r}\otimes m_s\otimes1^{\otimes t}) = 0,$$ for all $n>0$. Also, $m_0=0$. It follows from this composition identity that $m_1m_1 =0$ and cohomology makes sense.
2. The morphisms of $A_\infty$-algebras: if $A^*, B^*$ are $A_\infty$-algebras, a morphism $f\colon A^*\to B^*$ consists of $k$-linear morphisms $f_n\colon (A^*)^{\otimes n}\to B^*$ of degree $(1-n)$, such that the equation $$\sum_{r+s+t=n}(-1)^{r +st}f_{r+1+t}(1^{\otimes r}\otimes m_s\otimes1^{\otimes t}) = \sum_{i_1+\cdots+i_r=n}(-1)^\sigma m_r(f_{i_1}\otimes\cdots\otimes f_{i_r}),$$ where $$\sigma = \sum\limits_{j=1}^{r-1}(r-j)(i_j-1).$$ Setting $n=1$, one sees that $f_1$ is a chain map with respect to the differentials $m_1$ of $A^*$ and $B^*$, so we say that $f$ is a quasi-isomorphism of $A_\infty$-algebras if $f_1$ is a quasi-isomorphism of chain complexes.

My proof (or the attempt of one) goes as follows: first, as we are working over a field, treating $H^*(A^*)$ as a summand of $A^*$, we can find the projections $\pi^n\colon A^n\to H^n(A^*)$ for which we choose splittings $f_1^n\colon H^n(A^*)\to A^n$, and assume that $f^n_1\pi^n(a)=a$ for all $n$ and all $a\in H^n(A^*)$. Since $H^*(A^*)$ is cohomology wrt $m_1$, we may let $m_1'\colon H^*(A^*)\to H^*(A^*)$ be $0$. Then it is easy to see that $m_1^nf_1^n = 0 = f_1^{n+1}m_1'^n$ for all $n$, making $f_1$ a chain map. Furthermore, $f_1$ is a quasi-isomorphism.

Next, using $f_1$ we define $m_n'\colon H^*(A^*)^{\otimes n}\to A^*$ as the composite $$H^*(A^*)^{\otimes n}\xrightarrow{f_1^{\otimes n}} (A^*)^{\otimes n}\xrightarrow{m_n}A^*\xrightarrow{\pi}H^*(A^*)$$ for each $n\ge2$. Then the identities $$\sum_{r+s+t=n}(-1)^{r+st}m_{r+1+t}'(1^{\otimes r}\otimes m_s'\otimes1^{\otimes t}) = 0$$ hold, as $$m_{r+1+t}'(1^{\otimes r}m_s'\otimes1^{\otimes t}) = \pi m_{r+1+t}(f_1^{\otimes r}\otimes f_1\pi m_s(f_1^{\otimes s})\otimes f_1^{\otimes t}) = \pi m_{r+1+t}(1^{\otimes r}\otimes m_s\otimes1^{\otimes t})(f_1^{\otimes(r+s+t)}).$$ Thus, we have obtained an $A_\infty$-structure on $H^*(A^*)$, induced by that on $A^*$.

The next step is to find the morphisms $f_n\colon H^*(A^*)^{\otimes n}\to A^*$ for $n\ge2$. We start with the sum identity $$\sum_{r+s+t=n}(-1)^{r +st}f_{r+1+t}(1^{\otimes r}\otimes m_s\otimes1^{\otimes t}) = \sum_{i_1+\cdots+i_r=n}(-1)^\sigma m_r(f_{i_1}\otimes\cdots\otimes f_{i_r}).$$ On both sides (omitting the signs first), we consider the terms which contain $f_n$ or $f_{n+1}$. On the left hand side, we have $f_{n+1}(1^{\otimes n}\otimes m_0')$, $f_{n+1}(m_0'\otimes1^{\otimes n})$, $f_n(1^{\otimes(n-1)}\otimes m_1')$ and $f_n(m_1'\otimes1^{\otimes(n-1)})$, all of which are $0$ as $m_0'=0=m_1'$. On the right hand side, we have $m_1f_n$ whose sign is given by $(-1)^0$. Hence $$m_1f_n = \sum_{r+s+t=n}(-1)^{r +st}f_{r+1+t}(1^{\otimes r}\otimes m_s'\otimes1^{\otimes t}) - \sum_{i_1+\cdots+i_r = n, r\neq 1}(-1)^\sigma m_r(f_{i_1}\otimes\cdots\otimes f_{i_r}).$$ Denoting the right hand side of the previous equation by $g_n$, we then show that $m_1g_n = 0$ to make sure it lands in $Z^*(A^*)$, and then that $\pi g_n = 0$, as that implies it is a coboundary, and hence $g_n$ factors through $B^*(A^*) = \text{im}(m_1)$.

For the sake of an example, consider the case $n=2$. Here $$g_2= \sum_{r+s+t=2}(-1)^{r+st}f_{r+1+t}(1^{\otimes r}\otimes m_s'\otimes 1^{\otimes t}) - \sum_{i_1+\cdots+ i_r=2, r\neq1}(-1)^\sigma m_r(f_{i_1}\otimes\cdots\otimes f_{i_r}) = f_3(1\otimes 1\otimes m_0') - f_2(1\otimes m_1') + f_1m_2' - f_2(m_1'\otimes 1) + f_3(m_0'\otimes1\otimes1) - m_2(f_1\otimes f_1) = f_1m_2' - m_2(f_1\otimes f_1).$$ Then it follows from the identities $$m_1m_2 = m_2(1\otimes m_1 + m_1\otimes 1)$$ and $$m_1f_1=0$$ that $m_1g_2 = 0$, and furthermore $\pi g_2 = 0$, so $g_2$ lands in $B^*(A^*)$.

As we are dealing with vector spaces, we then obtain an $f_n\colon H^*(A^*)^{\otimes n}\to A^*$, such that $m_1f_n = g_n$. By construction, the $f\colon H^*(A^*)\to A^*)$ satisfies the required identities and is therefore a morphism of $A_\infty$-algebras. As $f_1$ is a quasi-isomorphism of chain complexes, $f$ is one of $A_\infty$-algebras.

I still need to prove the uniqueness part, but I'll do that if I know the stuff before it is even remotely correct. So please tell me if I did something dumb! Thanks for reading... Also, sorry for the potential typos! I tried to get rid of all but I may have missed something...